I remember a few years ago, BoJone provided a method to prove the mean value inequality (Arithmetic-Geometric Mean Inequality), but the proof was a bit long and somewhat dizzying (though the underlying idea was quite simple). Yesterday, during the "Mathematical Analysis" course, the teacher discussed this inequality and presented a proof using mathematical induction. However, it felt like it lacked that sense of concise and ingenious beauty. This reminded me of two clever proof methods I had studied before, but I spent all of yesterday trying to recall them without success. It wasn’t until today that I finally remembered them, so I am sharing them here for everyone and as a personal memorandum.
For several non-negative numbers x_i, we have: \frac{x_1+x_2+\dots+x_n}{n} \geq \sqrt[n]{x_1 x_2 \dots x_n}
This is denoted as A_n \geq G_n.
Proof 1: Mathematical Induction
This method is not necessarily simple, but it is very clever. The process of induction from n to n+1 is truly admirable. Proving Jensen’s inequality using mathematical induction follows the same recursive logic, and the AM-GM inequality is merely a special case of Jensen’s inequality.
Assume A_n \geq G_n holds, and we want to prove A_{n+1} \geq G_{n+1}. We have: \begin{aligned} 2n A_{n+1} &= (n+1)A_{n+1} + (n-1)A_{n+1} \\ &= [x_1 + x_2 + \dots + x_n] + [x_{n+1} + (n-1)A_{n+1}] \\ &\geq nG_n + n(x_{n+1} \cdot A_{n+1}^{n-1})^{\frac{1}{n}} \\ &\geq 2n(G_{n+1}^{n+1} \cdot A_{n+1}^{n-1})^{\frac{1}{2n}} \end{aligned}
Simplifying this yields: A_{n+1} \geq G_{n+1}.
This completes the induction from n to n+1. Other details are omitted.
Proof 2: Logarithmic Method
This method is even more clever and simpler; it might be the simplest
method I have ever seen. It utilizes a very basic formula: for all non-negative x, e^x \geq
1+x, and when x \geq -1, both
sides are non-negative.
(For ease of typesetting, we denote \exp(x) = e^x)
\begin{aligned} &\exp\left(\frac{n A_n}{G_n} - n\right) \\ =& \exp\left(\frac{x_1}{G_n}-1\right) \cdot \exp\left(\frac{x_2}{G_n}-1\right) \dots \exp\left(\frac{x_n}{G_n}-1\right) \\ \geq& \frac{x_1}{G_n} \cdot \frac{x_2}{G_n} \dots \frac{x_n}{G_n} \quad \text{(The exponents are clearly all $\geq -1$)} \\ =& 1 \end{aligned}
That is to say: \exp\left(\frac{n A_n}{G_n} - n\right) \geq 1
With a little simplification, we get A_n \geq G_n.
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